answer:The probability that heads will appear exactly k times, as is known, is C_{n}^{k} p^{k} q^{n-k}, where q=1-p is the probability of tails. Therefore, the event "Heads appeared an even number of times" has a probability ofP_{2}=C_{n}^{0} p^{0} q^{n}+C_{n}^{2} p^{2} q^{n-2}+C_{n}^{4} p^{4} q^{n-4}+ldots+C_{n}^{k} p^{k} q^{n-k}, where k is the largest even number not exceeding n.The complement of this event is the event "Heads appeared an odd number of times." The probability of this is obtained similarly by replacing even indices and powers with odd ones:P_{1}=C_{n}^{1} p^{1} q^{n-1}+C_{n}^{3} p^{3} q^{n-3}+C_{n}^{5} p^{5} q^{n-5}+ldots+C_{n}^{m} p^{m} q^{n-m}, where m is the largest odd number not exceeding n. The difference between these probabilities is: P_{2}-P_{1}=C_{n}^{0} p^{0} q^{n}-C_{n}^{1} p^{1} q^{n-1}+C_{n}^{2} p^{2} q^{n-2}-C_{n}^{3} p^{3} q^{n-3}+ldots pm C_{n}^{n} p^{n} q^{0}= (q-p)^{n}.On the other hand, P_{2}+P_{1}=1. From the resulting system of equations, we find:P_{2}=frac{1+(q-p)^{n}}{2}=frac{1+(1-2 p)^{n}}{2}In particular, for a symmetric coin p=0.5, so P_{2}=0.5.## Answera) 0.5; b) frac{1+(1-2 p)^{n}}{2}.Send a comment

answer:5. [pi-2,8-2 pi].Given that f(x) is an even function and strictly decreasing on the interval [0,1], we know that f(x) is strictly increasing on the interval [-1,0]. Combining this with the fact that f(x) has a period of 2, we know that [1,2] is a strictly increasing interval for f(x).Notice that,begin{array}{l}f(pi-2)=f(pi)=1, f(8-2 pi)=f(-2 pi)=f(2 pi)=2 .end{array}Then 1 leqslant f(x) leqslant 2Leftrightarrow f(pi-2) leqslant f(x) leqslant f(8-2 pi) text {. }Since 1<pi-2<8-2 pi<2, the original inequality holds if and only if x in[pi-2,8-2 pi].

answer:1. We start by identifying the term ( A_k = binom{1000}{k}(0.2)^k ) in the binomial expansion of ( (1+0.2)^{1000} ). We need to find the value of ( k ) for which ( A_k ) is the largest.2. Let ( n ) be the value of ( k ) such that ( A_n ) is the largest. This implies that ( A_n > A_{n-1} ) and ( A_n > A_{n+1} ). We can express these inequalities as: [ binom{1000}{n}(0.2)^n > binom{1000}{n-1}(0.2)^{n-1} ] [ binom{1000}{n}(0.2)^n > binom{1000}{n+1}(0.2)^{n+1} ]3. From the first inequality: [ binom{1000}{n}(0.2)^n > binom{1000}{n-1}(0.2)^{n-1} ] Simplifying, we get: [ frac{binom{1000}{n}}{binom{1000}{n-1}} > frac{1}{0.2} ] Using the property of binomial coefficients (binom{1000}{n} = frac{1000!}{n!(1000-n)!}) and (binom{1000}{n-1} = frac{1000!}{(n-1)!(1000-n+1)!}), we have: [ frac{frac{1000!}{n!(1000-n)!}}{frac{1000!}{(n-1)!(1000-n+1)!}} > 5 ] Simplifying further: [ frac{(n-1)!(1000-n+1)!}{n!(1000-n)!} > 5 ] [ frac{1000-n+1}{n} > 5 ] [ 1000 - n + 1 > 5n ] [ 1001 > 6n ] [ n < frac{1001}{6} ] [ n le leftlfloor frac{1001}{6} rightrfloor = 166 ]4. From the second inequality: [ binom{1000}{n}(0.2)^n > binom{1000}{n+1}(0.2)^{n+1} ] Simplifying, we get: [ frac{binom{1000}{n}}{binom{1000}{n+1}} > 0.2 ] Using the property of binomial coefficients (binom{1000}{n+1} = frac{1000!}{(n+1)!(1000-n-1)!}), we have: [ frac{frac{1000!}{n!(1000-n)!}}{frac{1000!}{(n+1)!(1000-n-1)!}} > 0.2 ] Simplifying further: [ frac{(n+1)!(1000-n-1)!}{n!(1000-n)!} > 0.2 ] [ frac{n+1}{1000-n} > 0.2 ] [ 5(n+1) > 1000 - n ] [ 6n > 995 ] [ n > frac{995}{6} ] [ n ge leftlceil frac{995}{6} rightrceil = 166 ]5. Combining the results from both inequalities, we have: [ 166 le n le 166 ] Therefore, ( n = 166 ).The final answer is ( boxed{166} ).

answer:A1 Since (-3)^{12}=3^{12}>0, it follows that -(-3)^{12}<0. The other numbers are positive.

answer:Given the function f(x) = ax^2 + bx + c (a > 0) satisfies f(1-x) = f(1+x) for any x in mathbb{R}, it can be deduced that the function is symmetric about x=1. Since a > 0, the function is decreasing in the interval (-infty, 1] and increasing in the interval [1, +infty). When x > 0, we have 3^x > 2^x > 1, thus f(3^x) > f(2^x). When x = 0, we have 3^x = 2^x = 1, thus f(3^x) = f(2^x). When x f(2^x). In summary, we have f(3^x) geq f(2^x). Therefore, the correct choice is boxed{text{A}}.Analysis: Given the function f(x) = ax^2 + bx + c (a > 0) satisfies f(1-x) = f(1+x) for any x in mathbb{R}, it can be deduced that the function is symmetric about x=1. Since a > 0, the function is decreasing in the interval (-infty, 1] and increasing in the interval [1, +infty). When x > 0, 3^x > 2^x > 1; when x = 0, 3^x = 2^x = 1; when x < 0, 3^x < 2^x < 1. This allows us to determine the relationship.

answer:Answer: 1 / 2 and 6 / 13, 4 / 13, 9 / 13, 6 / 13.Solution: Note that the only arrangement of four triangles so that they are similar to the larger one (and thus to each other) is such that each triangle has one vertex at a vertex of the larger triangle and the other two on its sides, with one point marked on each side of the larger triangle (A^{prime} on B C, B^{prime} on A C, C^{prime} on left.A Cright). Then there are exactly two options - angle C A^{prime} B^{prime}=angle C A B and angle C A^{prime} B^{prime}=angle C B A. The other angles are uniquely determined. Both options give a set of four similar triangles.Criteria: mp - only the first variant is considered.